A complete and practical solution book for the common school teacher . J-J-Hence, 400—133J = 266f sq. rd ABC. FIG. IS or 1334the 5 sq. of the triangle PROBLEM 408. A man takes for himself a central square of 40 acres, within asquare of 160 acres, and gives to his four sons rectangular farms equalin every respect and making up the remainder of the farm. The fatherand sons are to have dwellings centrally located in their respectivefarms: how far will each son be from his nearest brother, and from hisfather? Solution. (1) Let ABCD be the farm of 160 A., and LFXM the fath- ers farm. (2) Dr
A complete and practical solution book for the common school teacher . J-J-Hence, 400—133J = 266f sq. rd ABC. FIG. IS or 1334the 5 sq. of the triangle PROBLEM 408. A man takes for himself a central square of 40 acres, within asquare of 160 acres, and gives to his four sons rectangular farms equalin every respect and making up the remainder of the farm. The fatherand sons are to have dwellings centrally located in their respectivefarms: how far will each son be from his nearest brother, and from hisfather? Solution. (1) Let ABCD be the farm of 160 A., and LFXM the fath- ers farm. (2) Draw EL, MP, XN, FY, and ARXY, RBPM, PCEL and EDYF will be the sons farms. 206 FAIRCHILDS SOLUTION BOOK. (3)(4)(5)(6)(?) (8) (9) (10) (ID(12)(13) AB = BC=160rd. LF=FX=80rd. YF = EL = MP=40rd. YD=AR=120 rd. OK = ON+NK=40 20 —NX=60 rd.— 40rd. = 20 rd. Then, S0 = V (602+202)= rd. = the distanceeach sons house is fromthe fathers. ST=60 rd., and TZ=20 rd. /. SZ=60 rd.+20 rd. = 80 rd. ZS=40 rd. Then, SS=-i/(402+802)=8cson is from the FIG. 88. .44+ distance one PROBLEM 409. John has a square wheat field containing- 10 A., and hires Mart tocut it; he starts to cut at the southeast corner of the field: how far isthe machine from the southeast corner of the square after cutting- 5 A.,the machine being a 6 ft. cut? Solution. (1) Let ABCD be the wheat field, and EFGH thepart that remains aftercutting an integral num-ber of rounds. (2) KLMN=5 A., the part that remains providingthe machine cuts the first5 A. in an integral num-ber of rounds. (3) AD=AB = 40rd. = 660 ft. (4) NK=KL= ft. fig. 89. (5) FP=40rd.— rd. = ^2= ft. (6) ^6, the width of the machine = 16, or the integral number of rounds, and .49 rd. remaining. (7) .49 , ft., twice the width. (8) EF=KL+twice w=468 ft. (9) Area of EFGH=4682 =219024 sq. ft., and area of KLMN
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Keywords: ., bookcentury1800, bookdecade1890, booksubject, booksubjectgeometry
